Rd Sharma 2020 Solutions for Class 9 Math Chapter 14 Areas Of Parallelograms And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelograms And Triangles are extremely popular among Class 9 students for Math Areas Of Parallelograms And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 9 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 14.15:

Question 1:

If the given figure, ABCD is a parallelogram, AEDC and CFAD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.
 

Answer:

Given: Here in the question it is given

(1) ABCD is a parallelogram,

(2) and

(3) , AB = 16 cm

(4) AE = 8cm

(5) CF = 10cm

To Find : AD =?

Calculation: We know that formula for calculating the

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore,

Area of paralleogram ABCD = 16 × 8 ……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = AD × 10 ……(2)

Since equation 1 and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence fro equation (1) and (2),

This means that,

Hence we get the result as

Page No 14.15:

Question 2:

In Q.No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, find AB.

Answer:

Given: Here in the question it is given that

(1) ABCD is a parallelogram,

(2) and

(3)

(4) AD = 6 cm

(5) AE = 8cm

(6) CF = 10cm

To Find : AB =?

Calculation: We know that formula for calculating the

Area of paralleogram = base × height

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore, Area of paralleogram ABCd = 16 × 8

Area of Parallelogram ABCD = AB× 8 ……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = 6 × 10 ……(2)

Since equation 1and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence equation 1 is equal to equation 2

Which means that,

Hence we got the measure of AB equal to

 

Page No 14.15:

Question 3:

Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.

Answer:

Given: Here in the question it is given that

(1) Area of paralleogram ABCD = 124 cm2

(2) E is the midpoint of AB, which means

(3) F is the midpoint of CD, which means

To Find : Area of Parallelogram AEFD

Calculation: We know that formula for calculating the

Area of Parallelogram = base × height

Therefore,

Area of paralleogram ABCD = AB × AD (Taking base as AB and Height as AD ) ……(1)

Therefore,

Area of paralleogram AEFD = AE × AD (Taking base as AB and Height as AD ) ……(2)

()

= Area of Parallelogram ABCD (from equation1)

Hence we got the result Area of Parallelogram AEFD

 

Page No 14.15:

Question 4:

If ABCD is a parallelogram, then prove that

ar (ΔABD) = ar (ΔBCD) = ar (ΔABC) = ar (ΔACD) = 12 ar (||gm ABCD)

Answer:

Given: Here in the question it is given that

(1) ABCD is a Parallelogram

To Prove :

(1)

(2)

(3)

(4)

Construction: Draw

Calculation: We know that formula for calculating the

Area of Parallelogram = base × height

Area of paralleogram ABCD = BC × AE (Taking base as BC and Height as AE ……(1)

We know that formula for calculating the

Area of ΔADC = Base × Height

(AD is the base of ΔADC and AE is the height of ΔADC)

= Area of Parallelogram ABCD (from equation1)

Hence we get the result

Similarly we can show that 

(2)

(3)

(4)

 



Page No 14.3:

Question 1:

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

(i)


(ii)


(iii)


(iv)


(v)


(vi)

Answer:

GIVEN: Here in the question figure 1 to 6 are shown.

To Find :

(1) The figures which lie on the same base and between the same parallels.

(2) Write the common base and parallels.

As we know that ‘Two geometric figures are said to be on the same base and between the same parallels, if they have a common side(base) and the vertices (or vertex) opposite to the common base of each figure lie on a line parallel to the base.’

(1) ΔAPB and trapezium ABCD are on the same base CD and between the same parallels AB and CD.

(2) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

 

(3) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD 

 

(4) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD .

 

(5) ΔQRT and Parallelogram PQRS are on the same base QR and between the same parallels PS and QR.

 

(6) Parallelogram PQRS, AQRD, and BCQR are between the same parallels. Also Parallelogram PQRS, BPSC and APSD are between the same parallels.

 

 



Page No 14.44:

Question 1:

In the given figure, compute the area of quadrilateral ABCD.
 

Answer:

Given: Here from the given figure we get

(1) ABCD is a quadrilateral with base AB,

(2) ΔABD is a right angled triangle

(3) ΔBCD is a right angled triangle with base BC right angled at B

To Find: Area of quadrilateral ABCD

Calculation:

In right triangle ΔBCD, by using Pythagoreans theorem

 

.So

In right triangle ABD

Hence we get Area of quadrilateral ABCD =

Page No 14.44:

Question 2:

In the given figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of Δ OTS if PQ = 8 cm.

Answer:

Given: Here from the given figure we get

(1) PQRS is a square,

(2) T is the midpoint of PS which means

(3) U is the midpoint of PS which means

(4) QU = 8 cm

To find: Area of ΔOTS

Calculation:

Since it is given that PQ = 8 cm. So

Since T and U are the mid points of PS and QR respectively. So

Therefore area of triangle OTS is equals to

Hence we get the result that Area of triangle OTS is

 

Page No 14.44:

Question 3:

Compute the area of trapezium PQRS in the given figure.

Answer:

Given:

(1) PQRS is a trapezium in which SR||PQ..

(2) PT = 5 cm.

(3) QT = 8 cm.

(4) RQ = 17 cm.

To Calculate: Area of trapezium PQRS.

Calculation:

In triangle

.So

No area of rectangle PTRS

Therefore area of trapezium PQRS is

Hence the answer is

Page No 14.44:

Question 4:

In the given figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

Answer:

Given: In figure:

(1) ∠AOB = 90°

(2) AC = BC,

(3) OA = 012 cm,

(4) OC = 6.5 cm.

To find: Area of ΔAOB

Calculation:

It is given that AC = BC where C is the mid point of AB

We know that the mid point of hypotenuse of right triangle is equidistant from the vertices

Therefore

CA = BC = OC

CA = BC = 6.5

AB = 2 × 6.5 = 13 cm

Now inn triangle OAB use Pythagoras Theorem

So area of triangle OAB

Hence area of triangle is

 

Page No 14.44:

Question 5:

In the given figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Answer:

Given: Here from the given figure we get

(1) ABCD is a trapezium

(2) AB = 7 cm,

(3) AD = BC = 5 cm,

(4) DC = x cm

(5) Distance between AB and DC is 4 cm

To find:

(a) The value of x

(b) Area of trapezium

Construction: Draw AL CD, and BM CD

Calculation:

Since AL CD, and BM CD

Since distance between AB and CD is 4 cm. So

AL = BM = 4 cm, and LM = 7 cm

In triangle ADL use Pythagoras Theorem

Similarly in right triangle BMC use Pythagoras Theorem

Now

We know that,

We get the result as

Area of trapezium is

 



Page No 14.45:

Question 6:

In the given figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 25, find the area of the rectangle.

Answer:

Given: Here from the given figure we get

(1) OCDE is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) OE = 2√5cm

To find: Area of rectangle OCDE.

Calculation:

In right triangle ΔODE use Pythagoras Theorem

 

We know that,

Hence we get the result as area of Rectangle OCDE =

Page No 14.45:

Question 7:

In the given figure, ABCD is a trapezium in which AB || DC. Prove that
ar(Δ AOD) = ar(Δ BOC).
 

Answer:

Given:

ABCD is a trapezium with AB||DC

To prove: Area of ΔAOD = Area of ΔBOC

Proof:

We know that ‘triangles between the same base and between the same parallels have equal area’

Here ΔABC and ΔABD are between the same base and between the same parallels AB and DC.

Therefore

Hence it is proved that

 

Page No 14.45:

Question 8:

In the given figure, ABCD, ABFE and CDEF are parallelograms. Prove that
ar(Δ ADE) = ar(Δ BCF)

Answer:

Given:

(1) ABCD is a parallelogram,

(2) ABFE is a parallelogram

(3) CDEF is a parallelogram

To prove: Area of ΔADE = Area of ΔBCF

Proof:

We know that,” opposite sides of a parallelogram are equal”

Therefore for

Parallelogram ABCD, AD = BC

Parallelogram ABFE, AE = BF

Parallelogram CDEF, DE = CF.

Thus, in ΔADE and ΔBCF, we have

So be SSS criterion we have

This means that

Hence it is proved that

Page No 14.45:

Question 9:

In the given figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)

Answer:

Given:

(1) ABC and ABD are two triangles on the same base AB,

(2) CD bisect AB at O which means AO = OB

To Prove: Area of ΔABC = Area of ΔABD

Proof:

Here it is given that CD bisected by AB at O which means O is the midpoint of CD.

Therefore AO is the median of triangle ACD.

Since the median divides a triangle in two triangles of equal area

Therefore Area of ΔCAO = Area of ΔAOD ...... (1)

Similarly for Δ CBD, O is the midpoint of CD

Therefore BO is the median of triangle BCD.

Therefore Area of ΔCOB = Area of ΔBOD ...... (2)

Adding equation (1) and (2) we get

Area of ΔCAO + Area of ΔCOB = Area of ΔAOD + Area of ΔBOD

Area of ΔABC = Area of ΔABD

Hence it is proved that

Page No 14.45:

Question 10:

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).

Answer:

Given:

(1) ABC is a triangle

(2) AD is the median of ΔABC

(3) G is the midpoint of the median AD

To prove:

(a) Area of Δ ADB = Area of Δ ADC

(b) Area of Δ BGC = 2 Area of Δ AGC

Construction: Draw a line AM perpendicular to AC

Proof: Since AD is the median of ΔABC.

Therefore BD = DC

So multiplying by AM on both sides we get

In ΔBGC, GD is the median

Since the median divides a triangle in to two triangles of equal area. So

Area of ΔBDG = Area of ΔGCD

Area of ΔBGC = 2(Area of ΔBGD)

Similarly In ΔACD, CG is the median

Area of ΔAGC = Area of ΔGCD

From the above calculation we have

Area of ΔBGD = Area of ΔAGC

But Area of ΔBGC = 2(Area of ΔBGD)

So we have

Area of ΔBGC = 2(Area of ΔAGC)

Hence it is proved that

(1)

(2)

 

Page No 14.45:

Question 11:

A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).

Answer:

Given:

(1) ABC is a triangle

(2) D is a point on BC such that BD = 2DC

To prove: Area of ΔABD = 2 Area of ΔAGC

Proof:

In ΔABC, BD = 2DC

Let E is the midpoint of BD. Then,

BE = ED = DC

Since AE and AD are the medians of ΔABD and ΔAEC respectively

and

The median divides a triangle in to two triangles of equal area. So

Hence it is proved that

 



Page No 14.46:

Question 12:

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:

(i) ar (Δ ADO) = ar(Δ CDO)

(ii) ar (Δ ABP) = ar (Δ CBP).

Answer:

Given: Here from the given figure we get

(1) ABCD is a parallelogram

(2) BD and CA are the diagonals intersecting at O.

(3) P is any point on BO

To prove:

(a) Area of ΔADO = Area ofΔ CDO

(b) Area of ΔAPB = Area ofΔ CBP

Proof: We know that diagonals of a parallelogram bisect each other.

O is the midpoint of AC and BD.

Since medians divide the triangle into two equal areas

In ΔACD, DO is the median

Area of ΔADO = Area ofΔ CDO

Again O is the midpoint of AC.

In ΔAPC, OP is the median

Area of ΔAOP = Area of ΔCOP …… (1)

Similarly O is the midpoint of AC.

In ΔABC, OB is the median

Area of ΔAOB = Area of ΔCOB …… (2)

Subtracting (1) from (2) we get,

Area of ΔAOB − Area of ΔAOP = Area of ΔCOB − Area of ΔCOP

Area of ΔABP = Area of ΔCBP

Hence it is proved that

(a)

(b)

Page No 14.46:

Question 13:

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.

(i) Prove that ar (Δ ADF) = ar (Δ ECF)

(ii) If the area of Δ DFB = 3 cm2, find the area of ||gm ABCD.

Answer:

Given: Here from the given figure we get

(1) ABCD is a parallelogram with base AB,

(2) BC is produced to E such that CE = BC

(3) AE intersects CD at F

(4) Area of ΔDFB = 3 cm

To find:

(a) Area of ΔADF = Area of ΔECF

(b) Area of parallelogram ABCD

Proof: Δ ADF and ΔECF, we can see that

ADF = ECF (Alternate angles formed by parallel sides AD and CE)

AD = EC

DFA = CFA (Vertically opposite angles)

(ASA condition of congruence)

As

DF = CF

Since DF = CF. So BF is a median in ΔBCD

Since median divides the triangle in to two equal triangles. So

Since .So

Hence Area of parallelogram ABCD

Hence we get the result

(a)

(b)

 

Page No 14.46:

Question 14:

ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).

Answer:

Given:

(1) Diagonals AC and BD of a parallelogram ABCD intersect at point O.

(2) A line through O intersects AB at P point.

(3) A line through O intersects DC at Q point.

To find: Area of (ΔPOA) = Area of (ΔQOC)

Proof:

From ΔPOA and ΔQOC we get that

=

OA = OC

=

So, by ASA congruence criterion, we have

So

Area (ΔPOA) = Area (ΔQOC)

Hence it is proved that

Page No 14.46:

Question 15:

In the given figure, D and E are two points on BC such that BD = DE = EC. Show that a (Δ ABD) = ar (Δ ADE) = ar (Δ AEC).
 

Answer:

Given: In ΔABCD, D and E are two points on BC such that BD = DE = EC

To prove:

Proof: The ΔABD, ΔADE, and ΔAEC, are on the equal bases and their heights are equal

Therefore their areas are equal

So

Hence we get the result as

 

Page No 14.46:

Question 16:

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)

Answer:

Given:

(1) ABCD is a quadrilateral,

(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.

To prove: Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC

Construction: Draw AL perpendicular to BD and CM perpendicular to BD

Proof:

We know that

Area of triangle = × base× height

Area of ΔAPD = × DP × AL …… (1)

Area of ΔBPC = × CM × BP …… (2)

Area of ΔAPB = × BP × AL …… (3)

Area of ΔCPD = × CM × DP …… (4)

Therefore

Hence it is proved that

Page No 14.46:

Question 17:

If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

Answer:

Given: Here from the question we get

(1) ABCD is a parallelogram

(2) P is any point in the interior of parallelogram ABCD

To prove:

Construction: Draw DN perpendicular to AB and PM perpendicular AB

Proof: Area of triangle = × base× height

Area of ΔAPB = × AB × PM …… (1)

Also we know that: Area of parallelogram = base× height

Area of parallelogram ABCD = AB × DN …… (2)

Now PM < DN (Since P is a point inside the parallelogram ABCD)

Hence it is proved that

 

Page No 14.46:

Question 18:

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram AB CD.

Answer:

Given:

(1) ABCD is a parallelogram.

(2) E is a point on BA such that BE = 2EA

(3) F is a point on DC such that DF = 2FC.

To find:

Area of parallelogram

Proof: We have,

BE = 2EA and DF = 2FC

AB − AE = 2AE and DC − FC = 2FC

AB = 3AE and DC = 3FC

AE = AB and FC = DC

AE = FC [since AB = DC]

Thus, AE || FC such that AE = FC

Therefore AECF is a parallelogram.

Clearly, parallelograms ABCD and AECF have the same altitude and

AE = AB.

Therefore

Hence proved that

Page No 14.46:

Question 19:

In a Δ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:

(i) ar (Δ PBQ) = ar (Δ ARC)

(ii) ar (Δ PRQ) = 12 ar (Δ ARC)

(iii) ar (Δ RQC) = 38 ar (Δ ABC).

Answer:

Given:

(1) In a triangle ABC, P is the mid-point of AB.

(2) Q is mid-point of BC.

(3) R is mid-point of AP.

To prove:

(a) Area of ΔPBQ = Area of ΔARC

(b) Area of ΔPRQ = Area of ΔARC

(c) Area of ΔRQC = Area of ΔABC

Proof: We know that each median of a triangle divides it into two triangles of equal area.

(a) Since CR is a median of ΔCAP

Therefore …… (1)

Also, CP is a median of ΔCAB.

Therefore …… (2)

From equation (1) and (2), we get

Therefore …… (3)

PQ is a median of ΔABQ

Therefore

Since

Put this value in the above equation we get

…… (4)

From equation (3) and (4), we get

Therefore …… (5)

(b)

…… (6)

…… (7)

From equation (6) and (7)

…… (8)

From equation (7) and (8)

(c)

= …… (9)

 

Page No 14.46:

Question 20:

ABCD is a parallelogram, G is the point on  AB such that AG = 2 GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(i) ar (ADEG) = ar (GBCE)

(ii) ar (Δ EGB) = 16 ar (ABCD)

(iii) ar (Δ EFC) = 12 ar (Δ EBF)

(iv) ar (Δ EBG) = ar (Δ EFC)

(v) Find what portion of the area of parallelogram is the area of Δ EFG

Answer:

Given:

ABCD is a parallelogram

G is a point such that AG = 2GB

E is a point such that CE = 2DE

F is a point such that BF = 2FC

To prove:

(i)

(ii)

(iii)

(iv)

What portion of the area of parallelogram ABCD is the area of ΔEFG

Construction: draw a parallel line to AB through point F and a perpendicular line to AB through

PROOF:

(i) Since ABCD is a parallelogram,

So AB = CD and AD = BC

Consider the two trapeziums ADEG and GBCE:

Since AB = DC, EC = 2DE, AG = 2GB

, and

, and

So, and

Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal

Since

So

Hence

(ii) Since we know from above that

. So

Hence

(iii) Since height of triangle EFC and triangle EBF are equal. So

Hence

(iv) Consider the trapezium in which

(From (iii))

Now from (ii) part we have

(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x

Now consider the tow triangles CFI and CBH which are similar triangles

So by the property of similar triangle CI = k and IH = 2k

Now consider the triangle EGF in which

Now

(Multiply both sides by 2)

…… (2)

From (1) and (2) we have



Page No 14.47:

Question 21:

In the given figure, CD || AE and CY || BA.

(i) Name a triangle equal in area of ΔCBX

(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)

(iii) Prove that ar (BCZY) = ar (Δ EDZ).
 

Answer:

Given:

(1) CD||AE.

(2) CY||BA.

To find:

(i) Name a triangle equal in area of ΔCBX.

(ii) .

(iii) .

Proof:

(i) Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal. Therefore

Therefore area of triangle CBX is equal to area of triangle AXY

(ii) Triangle ADE and triangle ACE are on the same base AE and between the same parallels AE and CD.

(iii) Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have

Now we know that

 

Page No 14.47:

Question 22:

In the given figure, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (PQE) = ar (Δ CFD).

Answer:

Given:

(i) PSDA is a parallelogram.

(ii) .

(iii)

To find:

Proof:

Since AP||BQ||CR||DS and AD||PS

So PQ = CD …… (1)

In ΔBED, C is the mid point of BD and CF||BE

This implies that F is the mid point of ED. So

EF = FD …… (2)

In ΔPQE and ΔCFD, we have

PE = FD

, and [Alternate angles]

PQ = CD.

So, by SAS congruence criterion, we have

ΔPQE = ΔDCF

Hence proved that

 

Page No 14.47:

Question 23:

In the given figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = 911 ar (trap. (XYBA)

Answer:

Given: ABCD IS A trapezium in which

(a) AB||DC

(b) DC = 40 cm

(c) AB = 60 cm

(d) X is the midpoint of AD

(e) Y is the midpoint of BC

To prove:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii)

Construction: Join DY and produce it to meet AB produced at P.

Proof:

(i) In ΔBYP and ΔCYD

Y is the midpoint of BC also X is the midpoint of AD

Therefore XY||AP and

(ii) We have proved above that XY||AP

XY|| AP and AB||DC (Given in question)

XY|| DC

(iii) Since X and Y are the midpoints of AD and BC respectively.

Therefore DCYX and ABYX are of the same height say h cm.

 

Page No 14.47:

Question 24:

D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar (Δ BOE) = 18 ar (Δ ABC).

Answer:

Given: In ΔABC

(1) D is the midpoint of the side BC

(2) E is the midpoint of the side BD

(3) O is the midpoint of the side AE

To prove:

Proof: We know that the median of a triangle divides the triangle into two triangles of equal area.

Since AD and AE are the medians of ΔABC and ΔABD respectively. And OB is the median of ΔABE

…… (1)

…… (2)

…… (3)

Therefore

Hence we have proved that

 

Page No 14.47:

Question 25:

In the given figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (Δ ACQ).
 

Answer:

Given:

(1) X and Y are the, midpoints of AC and AB respectively.

(2) QP|| BC

(3) CYQ and BXP are straight lines.

To prove:

Proof: Since X and Y are the, midpoints of AC and AB respectively.

So XY||BC

ΔBYC and ΔBXC are on the same base BC and between the same parallels XY and BC.

Therefore

…… (1)

Similarly the quadrilaterals XYAP and XYQA are on the same base XY and between the same parallels XY and PQ. Therefore

…… (2)

Adding equation 1 and 2 we get

Hence we had proved that

 

Page No 14.47:

Question 26:

In the given figure, ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)

(iii) ar (Δ PEA) = ar (Δ QFD).

Answer:

Given: ABCD and AEFD are two parallelograms

To prove:

(i) PE = FQ

(ii)

(iii)

Proof: (i) and (iii)

In ΔAPE and ΔDQF

Therefore

, and

(ii) ΔPFA and ΔPFD are on the same base PF and between the same parallels PQ and AD.

From (1) and (2) we get

 



Page No 14.48:

Question 27:

In the given figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar (||gm DLOP) = ar (||gm BMOQ)

Answer:

Given:

(1) ABCD is a parallelogram

(2) O is any point of AC.

(3) PQ||AB and LM||AD

To prove:

Calculation:

We know that the diagonal of a parallelogram divides it into two triangles of equal area

Therefore we have

Since OC and AO are diagonals of parallelogram OQCL and AMOP respectively. Therefore

Subtracting (2) and (3) from (1) we get

Hence we get the result

 

Page No 14.48:

Question 28:

In a Δ ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:

(i) ar (Δ LCM) = ar (Δ LBM)

(ii) ar (Δ LBC) = (Δ MBC)

(iii) ar (Δ ABM) = ar (Δ ACL)

(iv) ar (Δ LOB) = ar (Δ MOC)

Answer:

Given:

In ΔABC, if L and M are points on AB and AC such that LM||BC

To prove:

(i)

(ii)

(iii)

(iv)

Proof: We know that triangles between the same base and between the same parallels are equal in area.

(i) Here we can see that ΔLMB and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (1)

(ii) Here we can see that ΔLBC and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (2)

(iii) From equation (1) we have,

(iv) From (2) we have,

 

Page No 14.48:

Question 29:

In the given figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (Δ BDE) = 14 ar (Δ ABC)

(ii) ar (Δ BDE) = 12 ar (Δ BAE)

(iii) ar (Δ BFE) =ar (Δ AFD)

(iv) ar (ΔABC) = 2 ar (Δ BEC)

(v) ar (Δ FED) = 18 ar (Δ AFC)

(vi) ar (Δ BFE) = 2 ar (EFD)

Answer:

Given:

(a) ΔABC and Δ BDE are two equilateral triangles

(b) D is the midpoint of BC

(c) AE intersect BC in F.

To prove:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Proof: Let AB = BC = CA = x cm.

Then BD = = DE = BE

(i) We have

(ii) We Know that ΔABC and ΔBED are equilateral triangles

BE||AC

(iii) We Know that ΔABC and ΔBED are equilateral triangles

AB || DE

(iv) Since ED is a median of Δ BEC

(v) We basically want to find out FD. Let FD = y

Since triangle BED and triangle DEA are on the same base and between same parallels ED and BE respectively. So

Since altitude of altitude of any equilateral triangle having side x is

So

…… (1)

Now

…… (2)

From (1) and (2) we get

(vi) Now we know y in terms of x. So

……. (3)

…… (4)

From (3) and (4) we get

 

Page No 14.48:

Question 30:

If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AXDE meets BC at Y.

Show that
(i) Δ MBC Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
 

Answer:

Given:

(1) ABCD is a right angled triangle at A

(2) BCED, ACFG and ABMN are the squares on the sides of BC, CA and AB respectively.

(3) , meets BC at Y.

To prove:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Proof:

(i)

…… (1)

(ii) Triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

Therefore

(Using (1)) …… (2)

(iii) Since ΔMBC and square MBAN are on the same base MB and between the same parallels MB and NC.

…… (3)

From (2) and (3) we get

(iv) In triangle FCB and ACE

…… (4)

(v) Since ΔACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX.

…… (5)

(vi) Since ΔFCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG

…… (6)

From (5) and (6) we get

(vii) Applying Pythagoras Theorem in ΔACB, WE get

 



Page No 14.59:

Question 1:

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 1

(d) 3 : 1

Answer:

Given: Two parallelogram with the same base and between the same parallels.

To find: Ratio of their area of two parallelogram with the same base and between the same parallels.

Calculation: We know that “Two parallelogram with the same base and between the same parallels, are equal in area”

Hence their ratio is ,

So the correct answer is,i.e. option (c).

 

Page No 14.59:

Question 2:

A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d) 1 : 3

Answer:

Given: A triangle and a parallelogram with the same base and between the same parallels.

To find: The ratio of the area of a triangle and a parallelogram with the same base and between the same parallels.

Calculation: We know that,” the area of a triangle is half the area of a parallelogram with the same base and between the same parallels.”

Hence the ratio of the area of a triangle and a parallelogram with the same base and between the same parallels is

Therefore the correct answer is option is (b).

 

Page No 14.59:

Question 3:

Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of Δ ABC. Then the area of ΔPQR is

(a) 12 sq. units

(b) 6 sq. units

(c) 4 sq. units

(d) 3 sq. units

Answer:

Given: (1) The Area of ΔABC = 24 sq units.

(2) ΔPQR is formed by joining the midpoints of ΔABC

To find: The area of ΔPQR

Calculation: In ΔABC, we have

Since Q and R are the midpoints of BC and AC respectively.

PQ || BA PQ || BP

Similarly, RQ || BP. So BQRP is a parallelogram.

Similarly APRQ and PQCR are parallelograms.

We know that diagonal of a parallelogram bisect the parallelogram into two triangles of equal area.

Now, PR is a diagonal of APQR.

∴ Area of ΔAPR = Area of ΔPQR ……(1)

Similarly,

PQ is a diagonal of PBQR

∴ Area of ΔPQR = Area of ΔPBQ ……(2)

QR is the diagonal of PQCR

∴ Area of ΔPQR = Area of ΔRCQ ……(3)

From (1), (2), (3) we have

Area of ΔAPR = Area of ΔPQR = Area of ΔPBQ = Area of ΔRCQ

But

Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC

4(Area of ΔPBQ) = Area of ΔABC

∴ Area of ΔPBQ

Hence the correct answer is option (b).

Page No 14.59:

Question 4:

The median of a triangle divides it into two

(a) congruent triangle

(b) isosceles triangles

(c) right triangles

(d) triangles of equal areas

Answer:

Given: A triangle with a median.

Calculation: We know that a ,”median of a triangle divides it into two triangles of equal area.”

Hence the correct answer is option (d).

 

Page No 14.59:

Question 5:

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =

(a) 4 cm2

(b) 8 cm2

(c) 12 cm2

(d) 10 cm2

Answer:

Given: In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of CA

(3) F is the midpoint of AB

(4) Area of ΔABC = 16 cm2

To find: The area of Trapezium FBCE

Calculation: Here we can see that in the given figure,

Area of trapezium FBCE = Area of ||gm FBDE + Area of ΔCDE

Since D and E are the midpoints of BC and AC respectively.

∴ DE || BA DE || BF

Similarly, FE || BD. So BDEF is a parallelogram.

Now, DF is a diagonal of ||gm BDEF.

∴ Area of ΔBDF = Area of ΔDEF ……(1)

Similarly,

DE is a diagonal of ||gm DCEF

∴ Area of ΔDCE = Area of ΔDEF ……(2)

FE is the diagonal of ||gm AFDE

∴ Area of ΔAFE = Area of ΔDEF ……(3)

From (1), (2), (3) we have

Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF

But

Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC

∴ 4 Area of ΔBDF = Area of ΔABC

Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm2 …….(4)

Now

Hence we get

Area of trapezium FBCE

There fore the correct answer is option (c).

Page No 14.59:

Question 6:

ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm2 and ar (ΔAPC) = 20 cm2, then ar (ΔAPB) =

(a) 15 cm2

(b) 20 cm2

(c) 35 cm2

(d) 30 cm2

Answer:

Given: (1) ABCD is a parallelogram

(2) P is any point on CD

(3) Area of ΔDPA = 15 cm2

(4) Area of ΔAPC = 20 cm2

To find: Area of ΔAPB

Calculation: We know that , “If a parallelogram and a a triangle are on the base between the same parallels, the area of triangle is equal to half the area of the parallelogram.”

Here , ΔAPB and ΔACB are on the same base and between the same parallels.

(since AC is the diagonal of parallelogram ABCD, diagonal of a parallelogram divides the parallelogram in two triangles of equal area)

Hence the correct answer is option (c).



Page No 14.60:

Question 7:

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

(a) 28 cm2

(b) 48 cm2

(c) 96 cm2

(d) 24 cm2

Answer:

Given: Rhombus with diagonals measuring 16cm and 12 cm.

To find: Area of the figure formed by lines joining the midpoints of the adjacent sides.

Calculation: We know that, ‘Area of a rhombus is half the product of their diagonals’.

H and F are the midpoints of AD and BC respectively.

Now ABCD is a parallelogram which means

……..(1)

……(2)

From 1 and 2 we get that ABFH is a parallelogram.

Since Parallelogram FHAB and ΔFHE are on the base FH and between the same parallels HF and AB.

……(3)

Similarly ,

……(4)

Adding 3 and 4 we get,

Hence the correct answer is option (b).

Page No 14.60:

Question 8:

A, B, C, D are mid-points of sides of parallelogram PQRS. If ar (PQRS) = 36 cm2, then ar (ABCD) =

(a) 24 cm2

(b) 18cm2

(c) 30 cm2

(d) 36 cm2

Answer:

Given:

(1) PQRS is a parallelogram.

(2) A, B, C, D are the midpoints of the adjacent sides of Parallelogram PQRS.

(3)

To find:

Calculation:

A and C are the midpoints of PS and QR respectively.

Now PQRS is a parallelogram which means

AP = CQ ……..(1)

Also, PS || QR

AP || CQ ……(2)

From 1 and 2 we get that APCQ is a parallelogram.

Since Parallelogram APCQ and ΔABC are on the base AC and between the same parallels AC and PQ.

……(3)

Similarly ,

……(4)

Adding 3 and 4 we get,

Hence the correct answer is option (b).

Page No 14.60:

Question 9:

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is

(a) a rhombus of area 24 cm2

(b) a rectangle of area 24 cm2

(c) a square of area 26 cm2

(d) a trapezium of area 14 cm2

Answer:

Given: Rectangle with sides 8 cm and 6cm

To find: Area of the figure which is formed by joining the midpoints of the adjacent sides of rectangle.

Calculation: Since we know that

For rhombus EFGH, EG is the one diagonal which is equal to DA

FH is the other diagonal which is equal to AB

Hence the result is option (a).

Page No 14.60:

Question 10:

If AD is median of ΔABC and P is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)

(a) 1 : 5

(b) 1 : 5

(c) 1 : 6

(d) 3 : 5

Answer:

Given: (1) AD is the Median of ΔABC

(2) P is a point on AC such that ar (ΔADP) : ar (ΔABD) =

To find: ar (ΔPDC) : ar (ΔABC)

We know that” the medians of the triangle divides the triangle in two two triangles of equal area.”

Since AD is the median of ΔABC,

ar (ΔABD) = ar (ΔADC) ……(1)

Also it is given that

ar (ΔADP) : ar (ΔABD) = ……(2)

Now,

Therefore,

Hence the correct answer is option (c).

Page No 14.60:

Question 11:

Medians of ΔABC intersect at G. If ar (ΔABC) = 27 cm2, then ar (ΔBGC) =

(a) 6 cm2

(b) 9 cm2

(c) 12 cm2

(d) 18 cm2

Answer:

Given: (1) Median of ΔABC meet at G.

(2) Area of ΔABC = 27 cm2

To find: Area of ΔBCG.

We know that the medians of the triangle divides each other in the ratio of 2:1

Hence,

Hence the correct answer is option (b).

Page No 14.60:

Question 12:

In a ΔABC if D and E are mid-points of BC and AD respectively such that ar (ΔAEC) = 4cm2, then ar (ΔBEC) =

(a) 4 cm2

(b) 6 cm2

(c) 8 cm2

(d) 12 cm2

Answer:

Given: In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of AD

(3) ar (ΔAEC) = 4 cm2

To find: ar (ΔBEC)

Calculation: We know that”the median of the triangle divides the triangle into two triangle of equal area”

Since AD is the median of ΔABC,

ar (ΔABD) = ar (ΔADC) …… (1)

EC is the median of ΔADC,

ar (ΔAEC) = ar (ΔDEC) …… (2)

⇒ ar (ΔDEC) = 4 cm2

EC is the median of ΔBED

ar (ΔBED) = ar (ΔDEC) …… (3)

From 2 and 3 we get,

ar (ΔBED) = ar (ΔAEC) …… (4)

⇒ ar (ΔBED) = 4 cm2

Now,

Hence the correct answer is option (c).

Page No 14.60:

Question 13:

In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
 

Answer:

Given: (1) ABCD is a parallelogram.

(2) AB = 12 cm

(3) AE = 7.5 cm

(4) CF = 15cm

To find: AD

Calculation: We know that,

Area of a parallelogram = base × height

Area of a parallelogram ABCD = DC ×AE (with DC as base and AE as height) ……(1)

Area of a parallelogram ABCD = AD ×CF (with DC as base and AE as height) ……(2)

Since equation 1 and 2 both are Area of a parallelogram ABCD

Hence the correct answer is option (b).

Page No 14.60:

Question 14:

In the given figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal Q is joined. The ratio ar (||gm XQRY) : ar (ΔQSR) =

(i) 1 : 4

(ii) 2 : 1

(iii) 1 : 2

(iv) 1 : 1

Answer:

Given: (1) PQRS is a parallelogram.

(2) X is the midpoint of PQ.

(3) Y is the midpoint of SR.

(4) SQ is the diagonal.

To find: Ratio of area of ||gm XQRY : area of ΔQRS.

Calculation: We know that the triangle and parallelogram on the same base and between the same parallels are equal in area.

∴ Ar (||gm PQRS) = Ar (ΔQRS)

(since X is the mid point of PQ and Y is the midpoint of SR)

Hence the correct answer is option (d).

Page No 14.60:

Question 15:

Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is

(a) ΔAOB

(b) ΔBOC

(c) ΔDOC

(d) ΔADC

Answer:

Given: (1) ABCD is a trapezium, with parallel sides AB and DC

(2) Diagonals AC and BD intersect at O

To find: Area of ΔAOD equals to ?

Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”

Therefore,

Hence the correct answer is option (b).

Page No 14.60:

Question 16:

ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

Answer:

Given: (1) ABCD is a trapezium, with parallel sides AB and DC

(2) Area of ΔADB = 24 cm2

(3) AB = 8 cm

To find: Height of ΔABC.

Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”

Here we can see that, ΔADB and ΔACB are on the same base AB.

Hence,

Area of ΔACB = Area of ΔADB

Area of ΔACB = 24

Hence the correct answer is option (c).



Page No 14.61:

Question 17:

ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is

(a) a : b

(b) (a + 3b): (3a + b)

(c) (3a + b) : (a + 3b)

(d) (2a + b) : (3a + b)

Answer:

Given: (1) ABCD is a trapezium, with parallel sides AB and DC

(2) AB = a cm

(3) DC = b cm

(4) E is the midpoint of non parallel sides AD.

(5) G is the midpoint of non parallel sides BC.

To find: Ratio of the area of the Quadrilaterals ABFE and EFCD.

Calculation: We know that, ‘Area of a trapezium is half the product of its height and the sum of the parallel sides.’

Since, E and F are mid points of AD and BC respectively, so h1 = h2

Area of trapezium ABFE

Now, Area (trap ABCD) = area (trap EFCD) + Area (ABFE)

Therefore,

Thus,

Hence, the correct option is (c)

 

Page No 14.61:

Question 18:

ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm2 and ar (ΔABOC) = 6 cm2, then area of rectangle ABCD is

(a) 9 cm2

(b) 12 cm2

(c) 15 cm2

(d) 18 cm2

Answer:

Given: A rectangle ABCD , O is a point in the interior of the rectangles such that

(1) ar (ΔAOB) = 3 cm2

(2) ar (ΔBOC) = 6 cm2

To find: ar (rect.ABCD)

Construction: Draw a line LM passing through O and parallel to AD and BC.

Calculation: We know that ,” If a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half the area of the parallelogram”

Here we can see that ΔAOD and rectangle AMLD are on the same base AD and between the same parallels AD and LM.

Hence ,

Similarly, we can see that ΔBOC and rectangle BCLM are on the same base BC and between the same parallels BC and LM

Hence,

We known that

Hence the correct answer is option (d).

Page No 14.61:

Question 19:

In the given figure, a parallelogram ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of ABCD = 12 perimeter of ABEF

Answer:

We know, opposite sides of a rectangle are equal.
Therefore, in rectangle ABEF,
AB = EF    ...(1)

Also, opposite sides of a parallelogram are equal.
Therefore, in parallelogram ABCD,
AB = DC    ...(3)

From (1) and (3),
DC = EF 
AB + EF = AB + DC    ...(5)

Now, we know that, of all the line segments, perpendicular segment is the shortest.
AF < AD
BE < BC
⇒ AF + BE < AD + BC    ...(6)

Adding (5) and (6), we get
AB + EF +  AF + BE < AB + DC + AD + BC
⇒ Perimeter of rectangle < perimeter of parallelogram

Hence, the correct option is (c).

Page No 14.61:

Question 20:

In the given figure, the area of parallelogram ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL

Answer:

We know, area of the parallelogram = Base × Altitude
Therefore,
The area of parallelogram ABCD = AB × DL
                                                      = DC × DL


Hence, the correct option is (c).

Page No 14.61:

Question 21:

ABCD is quadrilateral whose diagonal AC divided it into two parts, equal in area, then ABCD
(a) is a rectangle
(b) is always a rhombus
(c) is a parallelogram
(d) need not be any of (a), (b) or (c)

Answer:

The diagonal of a parallelogram, rectangle, rhombus or a square divides them into two parts, equal in area.

Therefore, if ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD need not be a rectangle, parallelogram or a rhombus. It can be a square as well.

Hence, the correct option is (d).

Page No 14.61:

Question 22:

The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

(a) ar (ΔABC)

(b) 12ar (ΔABC)

(c) 13ar (ΔABC)

(d) 14ar (ΔABC)

Answer:

Given: (1) ABCD is a triangle.

(2) mid points of the sides of ΔABC with any of the vertices forms a parallelogram.

To find: Area of the parallelogram

Calculation: We know that: Area of a parallelogram = base × height

Hence area of ||gm DECF = EC × EG

area of ||gm DECF = EC × EG

area of ||gm DECF = (E is the midpoint of BC)

area of ||gm DECF =

area of ||gm DECF =

Hence the result is option (b).

Page No 14.61:

Question 23:

In the given figure, ABCD and FECG are parallelograms equal in area. If ar (ΔAQE) = 12 cm2, then ar (||gm FGBQ) =

(a) 12 cm2

(b) 20 cm2

(c) 24 cm2

(d) 36 cm2
 

Answer:

Given: (1) Area of parallelogram ABCD is equal to Area of parallelogram FECG.

(2) If Area of ΔAQE is 12cm.

To find: Area of parallelogram FGBQ

Calculation: We know that diagonal of a parallelogram divides the parallelogram into two triangles of equal area.

 

It is given that,

Hence the correct answer is option (c).



Page No 14.62:

Question 1:

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is _________.

Answer:

Given:
A rhombus ABCD with diagonals 12 cm and 16 cm
i.e., AC = 16 cm and BD = 12 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.



Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
PQ || AC
PQ = 12AC
⇒ PQ = 
12(16)
⇒ PQ = 8 cm

In ∆ADC,
RS || AC
RS = 12AC
⇒ RS = 
12(16)
⇒ RS = 8 cm

In ∆BCD,
RQ || BD
RQ 12BD
⇒ RQ = 
12(12)
⇒ RQ = 6 cm

In ∆BAD,
SP || BD
SP 12BD
⇒ SP = 
12(12)
⇒ SP = 6 cm

Since, PQ = 8 cm = RS and RQ = 6 cm = SP
and Diagonals of a rhombus intersect at right angle.
angle between AC and BD is 90°
angle between PQ and QR is 90°
Therefore, PQRS is a rectangle
Thus, Area of rectangle = PQ × QR
                                       = 8 × 6
                                       = 48 cm2

Hence, the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is 48 cm2.

Page No 14.62:

Question 2:

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a ________ of area _________.

Answer:

Given: 
A rectangle ABCD of sides 8 cm and 6 cm
i.e., AB = 8 cm and AD = 6 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.



We can see that QS || AB and PR || AD
Also, QS = AB =  8 cm and PR = AD = 6 cm
Thus, angle between QS and PR is 90°
Therefore, PQRS is a rhombus.

Area of rhombus = 12× SQ × PR
                            = 
12× 8 × 6
                            = 4 × 6
                            = 24 cm2

Hence, the figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a rhombus of area 24 cm2.

Page No 14.62:

Question 3:

If P is any point on the median AD of a ΔABC, then ar ABPar ACP= _________.

Answer:

Given: P is any point on the median AD of a ΔABC

We know, median of a triangle divides it into two triangles of equal area.

Therefore, ar (ΔADB) = ar (ΔADC)       ...(1)
Also, ar (ΔPDB) = ar (ΔPDC)               ...(2)

Subtracting (2) from (1), we get
ar (ΔADB) − ar (ΔPDB) = ar (ΔADC) − ar (ΔPDC)
⇒ ar (ΔABP) = ar (ΔACP)
ar ABPar ACP= 1


Hence, ar ABPar ACP1.

Page No 14.62:

Question 4:

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is _________.

Answer:

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Area of triangle = 12 Area of the parallelogram
Area of triangleArea of parallelogram=12


Hence, if a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is 1 : 2.

Page No 14.62:

Question 5:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is _________.

Answer:

Parallelograms on equal bases and between the same parallels are equal in area.

Area of first parallelogram = Area of the second parallelogram
Area of first parallelogramArea of second parallelogram=11


Hence, the ratio of their areas is 1 : 1.

Page No 14.62:

Question 6:

ABCD is a parallelogram and X is the mid-point of AB. If ar(AXCD) = 24 cm2 , then ar(ΔABC) = ________.

Answer:

Given: 
ABCD is a parallelogram
X is the mid-point of AB
ar(AXCD) = 24 cm2

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔABC) 12ar(ABCD)     ...(1)

Since, X is the mid-point of AB
Therefore, ar(ΔXBC12ar(ABC)
                                   = 12×12ar(ABCD)    (From (1))
                                   = 14ar(ABCD)      ...(2)

Thus, ar(AXCD) = ar(ABCD) − ar(ΔXBC)
⇒ 24 = ar(ABCD) − 14ar(ABCD)         (From (2))
⇒ 24 = 34ar(ABCD)
⇒ ar(ABCD) = 24×43
⇒ ar(ABCD) = 8 × 4
⇒ ar(ABCD) = 32 cm2

From (1)
ar(ΔABC12ar(ABCD)
                 = 12× 32
                 = 16 cm2


Hence, ​ar(ΔABC) = 16 cm2.

Page No 14.62:

Question 7:

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar(ΔPAS) = _______.

Answer:

Given: 
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm
PS = 5 cm
A is any point on PQ


QS = radius of the circle = 13 cm     ...(1)

In ΔPQS
Using pythagoras theorem,
 QS2 = PS2 + PQ2
⇒ 132 = 52PQ2
⇒ PQ2 = 169 − 25
PQ2 = 144
⇒ PQ = 12 cm = SR            ...(2)


Thus,
ar(ΔRAS12× base × height
                 = 12× SR × PS
                 = 12× 12 × 5
                 = 30 cm2


Hence, ​ar(ΔRAS) = 30 cm2.

Disclaimer: The question is to find the area of ΔRAS instead of the area of ΔPAS.

Page No 14.62:

Question 8:

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC then ar(ΔABC) : ar(BDE) = _________.

Answer:

Given: 
ABC and BDE are two equilateral triangles
is the mid-point of BC


 arABC=34×side2               =34×BC2             ...1arBDE=34×side2               =34×BD2               =34×12BC2      D is the mid-point of BC               =34×14×BC2               =316×BC2             ...2Dividing 1 and 2, we getarABCarBDE=34BC2316BC2                 =41


Hence, ar(ΔABC) : ar(ΔBDE) = 4 : 1.
 

Page No 14.62:

Question 9:

If PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS, then ar(ΔASR) is _______ 90 cm2?

Answer:

Given: 
PQRS is a parallelogram with area 180 cm2
A is any point on the diagonal QS


 We know, the diagonal of the parallelogram bisects it into two triangles of equal area.
Thus, ar(ΔPQS) = ar(ΔQRS) = 12ar(PQRS)
⇒ ar(ΔPQS) = ar(ΔQRS) = 90 cm2

Therefore, ar(ΔASR) is always less than 90 cm2 unless or until the point A coincides with Q or S.

Hence, ar(ΔASR) is less than 90 cm2



Page No 14.63:

Question 10:

The mid points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to ________.

Answer:

Given: 
ABC is a triangle


Let D is the mid-point of ABE is the mid-point of BC and F is the mid-point of AC.

ADEF is a parallelogram having 2 triangles of equal area i.e., ∆ADF and ∆DEF.

But the ∆ABC is divided in 4 triangles of equal area i.e., ∆ADF, ∆DEF, ∆BED and ∆CEF.

Thus, area of ∆ABC = × area of the parallelogram ADEF.


Hence, the mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to half the area of the triangle ABC.

Page No 14.63:

Question 11:

A median of a triangle divides it into two ___________.

Answer:

Let ABC be a triangle with a mid-point D on BC.
Therefore, BD = DC
Let AE be the altitude from A on BC.

Now, ar(∆ABD) = 12× base × height
                          = 12× BD × AE

Also, ar(∆ACD) = 12× base × height
                          = 12× CD × AE
                          =
 12× BD × AE         (∵ BD = CD)
                          = ar(∆ABD


Hence, a median of a triangle divides it into two triangles of equal area.

Page No 14.63:

Question 12:

If ABCD is a rectangle, E and F are the mid-points of BC and AD respectively and G is any point on EF. Then ar(ΔGAB) : ar (rectangle ABCD) = _________.

Answer:

Given: 
ABCD is a rectangle
E and F are the mid-points of BC and AD respectively
is any point on EF


Since, E and F are the mid-points of BC and AD respectively.
Therefore, ar(BEFA) = ar(ECDF) = 12× ar(ABCD)            ...(1)

We know, if a triangle and a rectangle are on the same base and between the same parallels, then the area of the triangle is half the area of the rectangle.
Thus, ar(ΔGAB12ar(BEFA)             ...(2)

From (1) and (2),
ar(ΔGAB) = 12×12× ar(ABCD)
⇒ ar(ΔGAB) = 14× ar(ABCD)
⇒ arGABarrectangle ABCD=14


Hence, ar(ΔGAB) : ar(rectangle ABCD) = 1 : 4.

Page No 14.63:

Question 13:

PQRS is a square of side 8 cm. T and U and  respectively the mid points of PS and QR. If TU and QS intersect at 0, then ar(ΔOTS) = ________.


 

Answer:

Given: 
PQRS is a square of side 8 cm.
and are respectively the mid-points of PS and QR
TU and QS intersect at O



In ΔQOU and ΔOTS,
QOU = ∠TOS  (vertically opposite angles)
OQU = ∠OST  (alternate angles)
QU = TS (mid-points of sides of a square)

By AAS property,
ΔQOU ≅ ΔOTS

Thus, OU = OT (by CPCT)
⇒ OU + OT = PQ = 8 cm
⇒ OU = OT = 4 cm             ...(1)

Also, TS = 4 cm (T is the mid-point of PS)       ...(2)

ar(ΔOTS) = 12× base × height
                = 12× TS × OT
                = 12× 4 × 4    (From (1) and (2))
                = 8 cm2


Hence, ar(ΔOTS) = 8 cm2.

Page No 14.63:

Question 14:

In the given figure, ABCD and EFGD are two parallelograms and G is the mid point of CD. Then, ar(ΔDPC) : ar(EFGD) = ________.

Answer:

Given: 
ABCD and EFGD are two parallelograms.
G is the mid point of CD


Since, G is the mid point of CD
Therefore, DG = GC

Since, ΔDPG and ΔGPC have equal base and common height,
Thus, ar(ΔDPG) = ar(ΔGPC
⇒ ar(ΔDPC) = 2 ar(ΔDPG)       ...(1)

Also, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔDPG12ar(EFGD)     ...(2)

From (1) and (2),
ar(ΔDPC) = 2 × 12ar(EFGD)
⇒ ar(ΔDPC) = ar(EFGD)


Hence, ar(ΔDPC) : ar(EFGD) = 1 : 1.

Page No 14.63:

Question 15:

If the given figure, PQRS and EFRS are two parallelograms, then ar(||gm PQRS) : ar(ΔMFR) = ________.

Answer:

Given: 
PQRS and EFRS are two parallelograms



PQRS and EFRS are two parallelograms lying on the same base SR and between the same parallels SR and PF.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(PQRS) = ar(EFRS)             ...(1)

Also, ΔMFR and parallelogram EFRS is lying on the same base FR and between the same parallels SR and EF.

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Thus, ar(ΔMFR12ar(EFRS)     ...(2)

From (1) and (2),
ar(ΔMFR) = 12ar(PQRS)
⇒ ar(||gm PQRS) = 2 ar(ΔMFR)


Hence, ar(||gm PQRS) : ar(ΔMFR) = 2 : 1.

Page No 14.63:

Question 16:

In the given figure, if ABCD is a parallelogram of area 90 cm2. Then, ar(||gm ABEF) = ________ ar(ΔABD) = _______ and ar(ΔBEF) = _________.

Answer:

Given: 
ABCD is a parallelogram of area 90 cm2


ABCD and ABEF are two parallelograms lying on the same base AB and between the same parallels AB and CF.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(ABCD) = ar(ABEF) = 90 cm2           ...(1)

Also, ΔABD and parallelogram ABEF is lying on the same base AB and between the same parallels AF and BE.

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Thus, ar(ΔABD12ar(ABEF)
⇒ ar(||gm ABEF) = 2 ar(ΔABD)          ...(2)

From (1) and (2),
ar(||gm ABEF) = 2 ar(ΔABD) = 90 cm2

Also, diagonal of a parallelogram divides it into two triangles of equal area.
Thus,  ar(ΔBEF12ar(ABEF)
                            = 12(90)
                            = 45 cm2


Hence, ar(||gm ABEF) = 2 ar(ΔABD) = 90 cm2 and ar(ΔBEF) = 45 cm2.

Page No 14.63:

Question 1:

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar (ΔABC) : ar (ΔBDE).

Answer:

Given: (1) ΔABC is equilateral triangle.

(2) ΔBDE is equilateral triangle.

(3) D is the midpoint of BC.

To find:

PROOF : Let us draw the figure as per the instruction given in the question.

We know that area of equilateral triangle = , where a is the side of the triangle.

Let us assume that length of BC is a cm.

This means that length of BD is cm, Since D is the midpoint of BC.

------(1)

------(2)

Now, ar(ΔABC) : ar(ΔBDE) = (from 1 and 2)

=

Hence we get the result ar(ΔABC) : ar(ΔBDE) =



Page No 14.64:

Question 2:

In the given figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Answer:

Given: (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

To find: Area of rectangle CDEF.

Calculation: We know that,

Area of parallelogram = base × height

The Area of parallelogram and a rectangle on the same base and between the same parallels are equal in area.

Here we can see that rectangle ABCD and Parallelogram CDEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of Rectangle CDEF =

Page No 14.64:

Question 3:

In the given figure, find the area of ΔGEF.

Answer:

Given: (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

To find: Area of ΔGEF.

Calculation: We know that,

Area of Parallelogram = base × height

If a triangle and a parallelogram are on the same base and between the same parallels , the area of the triangle is equal to half of the parallelogram

Here we can see that Parallelogram ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as

Page No 14.64:

Question 4:

In the given figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ΔEFG.

Answer:

Given: (1) ABCD is a rectangle.

(2) AB = 10 cm

(3) AD = 5cm

To find: Area of ΔEGF.

Calculation: We know that,

Area of Rectangle = base × height

If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram

Here we can see that Rectangle ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of ΔGEF =

Page No 14.64:

Question 5:

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar (ΔRAS)

Answer:

Given: Here from the given figure we get

(1) PQRS is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) PS = 5cm

(3) PR = 13cm(radius of the quadrant)

To find: Area of ΔRAS.

Calculation: In right ΔPSR, (Using Pythagoras Theorem)

Hence we get the Area of ΔRAS =

Page No 14.64:

Question 6:

In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8cm and PQ and BD intersect at O, then find area of ΔOPB.

Answer:

Given: Here from the given question we get

(1) ABCD is a square,

(2) P is the midpoint of AB

(3) Q is the midpoint of CD

(4) PQ and BD intersect at O.

(5) AB = 8cm

To find : Area of ΔOPB

Calculation: Since P is the midpoint of AB,

BP = 4cm ……(1)

Hence we get the Area of ΔOBP =

 

Page No 14.64:

Question 7:

ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF.

Answer:

Given: Here from the given question we get

(1) ABC is a triangle

(2) D is the midpoint of BC

(3) E is the midpoint of CD

(4) F is the midpoint of A

Area of ΔABC = 16 cm2

To find : Area of ΔDEF

Calculation: We know that ,

The median divides a triangle in two triangles of equal area.

For ΔABC, AD is the median

For ΔADC , AE is the median .

Similarly, For ΔAED , DF is the median .

Hence we get Area of ΔDEF =

Page No 14.64:

Question 8:

PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ΔPBQ is 17cm2, find the area of ΔASR.

Answer:

Given: Here from the given figure we get

(1) PQRS is a trapezium having PS||QR

(2) A is any point on PQ

(3) B is any point on SR

(4) AB||QR

(5) Area of ΔBPQ = 17 cm2

To find : Area of ΔASR.

Calculation: We know that ‘If a triangle and a parallelogram are on the same base and the same parallels, the area of the triangle is equal to half the area of the parallelogram’

Here we can see that:

Area (ΔAPB) = Area (ΔABS) …… (1)

And, Area (ΔAQR) = Area (ΔABR) …… (2)

Therefore,

Area (ΔASR) = Area (ΔABS) + Area (ΔABR)

From equation (1) and (2), we have,

Area (ΔASR) = Area (ΔAPB) + Area (ΔAQR)

Area (ΔASR) = Area (ΔBPQ) = 17 cm2

Hence, the area of the triangle ΔASR is 17 cm2.

Page No 14.64:

Question 9:

ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3.1. If ar (ΔPBQ) = 10cm2, find the area of parallelogram ABCD.

Answer:

It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm2

Let CQ = x and QP = 3x

We need to find area of the parallelogram ABCD.

From the figure,

Area (PBQ) =

And,

Area (BQC) =

Now, let H be the perpendicular distance between AP and CD. Therefore,

Area (PCB) = …… (1)

Thus the area of the parallelogram ABCD is,

Area (ABCD) = AB × H

Area (ABCD) = 2BP × H

From equation (1), we get

Area (ABCD) = 4 × 30 = 120 cm2

Hence, the area of the parallelogram ABCD is 120 cm2.

Page No 14.64:

Question 10:

P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar (ΔABC) = 12 cm2, then find area of ΔEPC.

Answer:

Given: Area (ABC) = 12 cm2, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.

Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,

Area (APE) = Area (APD)

Area (APM) + Area (AME) = Area (APM) + Area (PMD)

Area (AME) = Area (PMD) …… (1)

Since, median divide triangles into two equal parts. So,

Area (ADC) = Area (ABC) = = 6 cm2

Area (ADC) = Area (MDCE) + Area (AME)

Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1))

Area (ADC) = Area (PEC)

Therefore,

Area (PEC) = 6 cm2.



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